# Class 10 RD Sharma Solutions – Chapter 2 Polynomials – Exercise 2.3

### Question 1. Apply division algorithm to find the quotient q(x) and remainder r(x) on dividing f(x) by g(x) in each of the following:

**(i) f(x) = x ^{3} – 6x^{2} + 11x – 6, g(x) = x^{2} + x + 1**

**(ii) f(x) = 10x ^{4} + 17x^{3} – 62x^{2} + 30x – 105, g(x) = 2x^{2} + 7x + 1**

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**(iii) f(x) = 4x ^{3} + 8x^{2} + 8x + 7, g(x) = 2x^{2} – x + 1**

**(iv) f(x) = 15x ^{3} – 20x^{2} + 13x – 12, g(x) = x^{2} – 2x + 2**

**Solution: **

(i)Here we have to divide f(x) = x^{3}– 6x^{2}+ 11x – 6 by g(x) = x^{2 }+ x + 1So, to get quotient q(x) and remainder r(x), we use division algorithm

Therefore,

Remainder r(x) = 17x – 1

Quotient q(x) = x – 7

(ii)Here we have to divide f(x) = 10x^{4 }+ 17x^{3}– 62x^{2}+ 30x – 105 by g(x) = 2x^{2}+ 7x + 1So, to get quotient q(x) and remainder r(x), we use division algorithm

Therefore,

Remainder r(x) = 53x – 1

Quotient q(x) = 5x

^{2}– 9x – 2

(iii)Here we have to divide f(x) = 4x^{3}+ 8x^{2}+ 8x + 7 by g(x) =2x^{2}– x + 1So, to get quotient q(x) and remainder r(x), we use division algorithm

Therefore,

Remainder r(x) = 11x + 2

Quotient q(x) = 2x – 5

(iv)f(x) = 15x^{3}– 20x^{2}+ 13x – 12, g(x) = x^{2 }– 2x + 2Here we have to divide f(x) = 15x

^{3}– 20x^{2}+ 13x – 12 by g(x) = x^{2 }– 2x + 2So, to get quotient q(x) and remainder r(x), we use division algorithm

Therefore,

Remainder r(x) = 3x + 32

Quotient q(x) = 15x + 10

### Question 2. Check whether the first polynomial is a factor of the second polynomial by applying the division algorithm:

**(i) g(t) = t ^{2} – 3; f(t) = 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12**

**(ii) g(x) = x ^{2} – 3x + 1; f(x) = x^{5} – 4x^{3} + x^{2} + 3x + 1**

**(iii) g(x) = 2x ^{2} – x + 3; f(x) = 6x^{5} − x^{4} + 4x^{3} – 5x^{2} – x – 15**

**Solution:**

(i)Here, we have to check whether g(t) = t^{2}– 3 is a factor of f(t) = 2t^{4}+ 3t^{3}– 2t^{2}– 9t – 12So, by using division algorithm, we get

As the remainder left is 0.

Therefore,

g(t) = t

^{2}– 3 is a factor of f(t) = 2t^{4}+ 3t^{3}– 2t^{2}– 9t – 12

(ii)Here, we have to check whether g(x) = x^{2 }– 3x + 1 is a factor off(x) = x^{5}– 4x^{3}+ x^{2}+ 3x + 1So, by using division algorithm, we get

As the remainder left is 2.

Therefore,

g(x) = x

^{2 }– 3x + 1 is not a factor of f(x) = x^{5}– 4x^{3}+ x^{2}+ 3x + 1

(iii)Here, we have to check whether g(x) = 2x^{2}– x + 3 is a factor of f(x) = 6x^{5}− x^{4}+ 4x^{3}– 5x^{2}– x – 15So, by using division algorithm, we get

As the remainder left is 0.

Therefore,

g(x) = 2x

^{2}– x + 3 is a factor of f(x) = 6x^{5}− x^{4}+ 4x^{3}– 5x^{2}– x – 15

### Question 3. Obtain all zeroes of the polynomial f(x) = f(x) = 2x^{4} + x^{3} – 14x^{2} – 19x – 6, if two of its zeroes are -2 and -1.

**Solution:**

Given: f(x) = 2x

^{4 }+ x^{3}– 14x^{2}– 19x – 6Here we have given the two zeroes of the polynomial that are -2 and -1,

Hence, its factors will be (x + 2) and (x + 1)

Further,

(x + 2)(x + 1) = x

^{2}+ x + 2x + 2 = x^{2}+ 3x + 2So, by using division algorithm, we get

f(x) = 2x

^{4}+ x^{3}– 14x^{2}– 19x – 6 = (2x^{2}– 5x – 3)(x^{2 }+ 3x + 2)= (2x + 1)(x – 3)(x + 2)(x + 1)

Hence, the factors of f(x) = 2x

^{4 }+ x^{3}– 14x^{2}– 19x – 6 are (2x + 1), (x – 3), (x + 2), (x + 1)Therefore, the zeroes of the polynomial are -1/2, 3, -2, -1

### Question 4. Obtain all zeroes of f(x) = x^{3} + 13x^{2} + 32x + 20, if one of its zeroes is -2.

**Solution:**

We have been given the zero of the polynomial f(x) = x

^{3}+ 13x^{2}+ 32x + 20 is -2.Hence, its factor is (x + 2).

So, by using division algorithm, we get

Thus,

f(x) = x

^{3}+ 13x^{2}+ 32x + 20= (x

^{2}+ 11x + 10)(x + 2)= (x

^{2 }+ 10x + x + 10)(x + 2)= (x + 10)(x + 1)(x + 2)

Hence, the factors of f(x) = x

^{3}+ 13x^{2}+ 32x + 20 are (x + 10), (x + 1), (x + 2)Thus, the zeroes of the polynomial are -1, -10, -2.

### Question 5. Obtain all zeroes of the polynomial f(x) = x^{4 }– 3x^{3} – x^{2} + 9x – 6, if the two of its zeroes are -√3 and √3.

**Solution:**

Here, we are given two zeros of the polynomial f(x) = x

^{4 }– 3x^{3}– x^{2}+ 9x – 6 that are -√3 and √3.Thus, the factors are (x + √3)(x − √3) ⇒ x

^{2}– 3.So, by using division algorithm, we get

Hence,

f(x) = x

^{4}– 3x^{2}– x^{2 }+ 9x – 6 = (x^{2 }– 3)(x^{2}– 3x + 2)(x + √3)(x – √3)(x

^{2 }– 2x – 2 + 2)= (x + √3)(x – √3)(x – 1)(x – 2)

Thus, the factors of f(x) = x

^{4}– 3x^{3}– x^{2}+ 9x – 6 are (x + √3)(x – √3)(x – 1)(x – 2).Therefore, the zeroes of the polynomial are -√3, √3, 1, 2.

### Question 6. Obtain all zeroes of the polynomial f(x) = 2x^{4} – 2x^{3} – 7x^{2} + x – 1, if the two of its zeroes are -√3/2 and √3/2.

**Solution:**

Here, we are given two zeros of the polynomial f(x) = 2x

^{4}– 2x^{3}– 7x^{2}+ x – 1 that are -√3/2 and √3/2.Thus, the factors are ⇒ x

^{2}– 3/2.So, by using division algorithm, we get

Hence,

Factors of f(x) = 2x

^{4}– 2x^{3}– 7x^{2}+ x – 1 are .Thus, the zeroes of the polynomial are -1, 2, -√3/2 and √3/2.

### Question 7. Find all the zeroes of the polynomial x^{4} + x^{3 }– 34x^{2} – 4x + 120, if the two of its zeroes are 2 and – 2.

**Solution:**

Here, we are given two zeros of the polynomial x

^{4}+ x^{3}– 34x^{2}– 4x + 120 that are 2 and -2.Thus, the factors are (x + 2)(x – 2)⇒ x

^{2}– 4.So, by using division algorithm, we get

Hence,

x

^{4}+ x^{3}– 34x^{2}– 4x + 120 = (x^{2}– 4)(x^{2}+ x – 30)= (x – 2)(x + 2)(x

^{2}+ 6x – 5x – 30)= (x – 2)(x + 2)(x + 6)(x – 5)

So, the factors of x

^{4}+ x^{3}– 34x^{2}– 4x + 120 are (x – 2), (x + 2), (x + 6), (x – 5)Thus, the zeroes of the polynomial = x = 2, – 2, – 6, 5

### Question 8. Find all the zeroes of the polynomial 2x^{4} + 7x^{3} – 19x^{2} – 14x + 30, if the two of its zeroes are √2 and -√2.

**Solution:**

Here, we are given two zeros of the polynomial 2x

^{4}+ 7x^{3}– 19x^{2}– 14x + 30^{2}that are √2 and -√2.Thus, the factors are (x + √2)(x – √2) ⇒ x

^{2}– 2.So, by using division algorithm, we get

Hence,

2x

^{4}+ 7x^{3}– 19x^{2}– 14x + 30 = (x^{2}– 2)(2x^{2}+ 7x – 15)= (2x

^{2 }+ 10x – 3x – 15)(x + √2)(x – √2)= (2x – 3)(x + 5)(x + √2)(x – √2)

So, the factors of 2x

^{4}+ 7x^{3}– 19x^{2}– 14x + 30 are (2x – 3), (x + 5), (x + √2), (x – √2)Thus, the zeroes of the polynomial is √2, -√2, -5, 3/2.

### Question 9. Find all the zeroes of the polynomial f(x) = 2x^{3} + x^{2} – 6x – 3, if two of its zeroes are -√3 and √3.

**Solution: **

Here, we are given two zeros of the polynomial f(x) = 2x

^{3}+ x^{2}– 6x – 3 that are -√3 and √3.Thus, the factors are (x + √3)(x – √3) ⇒ x

^{2}– 3.So, by using division algorithm, we get

Hence,

f(x) = 2x

^{3}+ x^{2}– 6x – 3= (x

^{2}– 3)(2x + 1)= (x + √3)(x – √3)(2x + 1)

Factors of f(x) = 2x

^{3}+ x^{2}– 6x – 3 are (x + √3), (x – √3), 2x + 1Thus, the zeroes for the given polynomial are √3, -√3, -1/2

### Question 10. Find all the zeroes of the polynomial f(x) = x^{3} + 3x^{2} – 2x – 6, if the two of its zeroes are √2 and -√2.

**Solution:**

Here, we are given two zeros of the polynomial f(x) = x

^{3 }+ 3x^{2}– 2x – 6 that are √2 and -√2.Thus, the factors are (x + √2)(x – √2)⇒ x

^{2}– 2.So, by using division algorithm, we get

Hence,

f(x) = x

^{3}+ 3x^{2}– 2x – 6= (x

^{2}– 2)(x + 3)= (x + √2)(x – √2)(x + 3)

Factors of f(x) = x

^{3}+ 3x^{2}– 2x – 6 are (x + √2), (x – √2), (x + 3)Thus, the zeroes of the given polynomial is -√2, √2, and – 3.

### Question 11. What must be added to the polynomial f(x) = x^{4} + 2x^{3} – 2x^{2} + x − 1 so that the resulting polynomial is exactly divisible by g(x) = x^{2} + 2x − 3.

**Solution:**

Here we have to add to the polynomial f(x) = x

^{4}+ 2x^{3}– 2x^{2}+ x − 1 so that theresulting polynomial is exactly divisible by g(x) = x

^{2}+ 2x − 3.So, divide f(x) = x

^{4}+ 2x^{3}– 2x^{2}+ x − 1 by g(x) = x^{2}+ 2x − 3 to get the answer.As the remainder left is (x – 2) to get the resulting polynomial exactly divisible by

g(x) = x

^{2 }+ 2x − 3 we must add (x – 2) to f(x) = x^{4}+ 2x^{3}– 2x^{2}+ x − 1.

### Question 12. What must be subtracted from the polynomial f(x) = x^{4} + 2x^{3} – 13x^{2 }–12x + 21 so that the resulting polynomial is exactly divisible by g(x) = x^{2 }– 4x + 3.

**Solution:**

Here we have to subtract to the polynomial f(x) = x

^{4}+ 2x^{3}– 13x^{2}– 12x + 21so that the resulting polynomial is exactly divisible by g(x) = x

^{2}– 4x + 3.So, divide f(x) = x

^{4}+ 2x^{3}– 13x^{2}– 12x + 21 by g(x) = x^{2}– 4x + 3 to get the answer.As the remainder left is (2x – 3) to get the resulting polynomial exactly divisible by

g(x) = x

^{2}– 4x + 3 we must add (2x – 3) to f(x) = x^{4}+ 2x^{3}– 13x^{2}– 12x + 21.

### Question 13. Given that √2 is a zero of the cubic polynomial f(x) = 6x^{3 }+ √2x^{2}– 10x – 4√2, find its other two zeroes.

**Solution:**

Here, we are given that √2 is the zero of the cubic polynomial

f(x) = 6x

^{3 }+ √2x^{2}– 10x – 4√2, thus, factor of the polynomial is (x – √2)So, by using division algorithm, we get

Hence,

f(x) = 6x

^{3 }+ √2x^{2 }– 10x – 4√2= (x – √2)(6x

^{2 }+ 7√2x + 4)= (x – √2)(6x

^{2 }+ 4√2x + 3√2x + 4)= (x – √2)(3x + 2√2)(2x + √2)

The factors of f(x) = 6x

^{3 }+ √2x^{2}– 10x – 4√2 are (x – √2), (3x + 2√2), (2x + √2)Therefore, the zeros of the polynomial are -2√2/3, -√2/2, √2

### Question 14. Given that x – √5 is a factor of the cubic polynomial x^{3 }– 3√5x^{2 }+ 13x – 3√5, find all the zeroes of the polynomial.

**Solution: **

Here, we have x – √5 as factor of the cubic polynomial x

^{3 }– 3√5x^{2 }+ 13x – 3√5To find all the zeros of the polynomial, we have to divide the polynomial x

^{3 }– 3√5x^{2 }+ 13x – 3√5 by the factor x – √5Hence,

x

^{3 }– 3√5x^{2 }+ 13x – 3√5= (x – √5)(x

^{2 }– 2√5 + 3)= (x – √5)(x – (√5 + √2))(x – (√5 – √2))

So, the factors of the cubic polynomial x

^{3 }– 3√5x^{2 }+ 13x – 3√5 are (x – √5), (x – (√5 + √2)), (x – (√5 – √2))Therefore, the zero of the polynomial are √5, (√5 – √2), (√5 + √2)